Given a singly linked list, determine if it is a palindrome. Follow up: Could you do it in O(n) time and O(1) space? 这道题让我们判断一个链表是否为回文链表,LeetCode中关于回文串的题共有六道,除了这道,其他的五道为Palindrome Number 验证回文数字,Validate Palindrome 验证回文字符串,Palindrome Partitioning 拆分回文…

Given a singly linked list, determine if it is a palindrome. Follow up:Could you do it in O(n) time and O(1) space? 题目意思: 给定一个单链表,判断它是不是回文串 进一步思考: 你可以在O(n)时间复杂度和O(1)空间复杂度完成吗? 解题思路: 方法一:通过反转链表实现 (1)使用快慢指针寻找链表中点 (2)将链表的后半部分就地逆置 (3)比较前后两半的元素是否一致 (4)恢复原始…

Description: Given a singly linked list, determine if it is a palindrome. Follow up:Could you do it in O(n) time and O(1) space? 判断一个单向链表是不是回文. 思路: 最简单的思路就是,用一个遍历一遍,存下来,再遍历反着做比较.时间复杂度O(n),空间复杂度O(n).竟然能过...... /** * Definition for singly-linked list.…

9 - Palindrome Number Determine whether an integer is a palindrome. Do this without extra space. Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using extra…

234. Palindrome Linked List 1. 使用快慢指针找中点的原理是fast和slow两个指针,每次快指针走两步,慢指针走一步,等快指针走完时,慢指针的位置就是中点.如果是偶数个数,正好是一半一半,如果是奇数个数,慢指针正好在中间位置,判断回文的时候不需要比较该位置数据. 注意好好理解快慢指针的算法原理及应用. 2. 每次慢指针走一步,都把值存入栈中,等到达中点时,链表的前半段都存入栈中了,由于栈的后进先出的性质,就可以和后半段链表按照回文对应的顺序比较. solution…

Palindrome Linked List Given a singly linked list, determine if it is a palindrome. Follow up:Could you do it in O(n) time and O(1) space? 解法一: 一次遍历,装入vector,然后再一次遍历判断回文. 时间复杂度O(n),空间复杂度O(n) /** * Definition for singly-linked list. * struct ListNode…

Question 234. Palindrome Linked List Solution 题目大意:给一个链表,判断是该链表中的元素组成的串是否回文 思路:遍历链表添加到一个list中,再遍历list的一半判断对称元素是否相等,注意一点list中的元素是Integer在做比较的时候要用equals,因为int在[-128,127)之间在内存中是存储在运行时常量池中,超出这个范围作为对象存储在堆中,==比较的是字面值和对象地址 Java实现: public boolean isPalindrom…

Reverse a singly linked list. click to show more hints. Hint: A linked list can be reversed either iteratively or recursively. Could you implement both? 之前做到Reverse Linked List II 倒置链表之二的时候我还纳闷怎么只有二没有一呢,原来真是忘了啊,现在才加上,这道题跟之前那道比起来简单不少,题目为了增加些许难度,让我们分别用…

Determine whether an integer is a palindrome. Do this without extra space. click to show spoilers. Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using ext…

2.7 Implement a function to check if a linked list is a palindrome. LeetCode上的原题,参见我之前的博客Palindrome Linked List 回文链表.…

234. Palindrome Linked List Easy Given a singly linked list, determine if it is a palindrome. Example 1: Input: 1->2 Output: false Example 2: Input: 1->2->2->1 Output: true Follow up:Could you do it in O(n) time and O(1) space? package leetcod…

Reverse a singly linked list. 这题因为palindrome linked list 的时候需要就顺便做了一下.利用三个指针:prev, now, next 相互倒腾就行. /** * Definition for singly-linked list. * function ListNode(val) { * this.val = val; * this.next = null; * } */ /** * @param {ListNode} head * @retu…

LeetCode:Palindrome Partitioning 题目如下:(把一个字符串划分成几个回文子串,枚举所有可能的划分) Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. For example, given s = "aab",Return [ […

LeetCode: Palindrome Partition Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. For example, given s = "aab", Return [ ["aa","b"], [&q…

Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s = "aab",Return 1 since the palindrome partitioning ["aa",…

LeetCode: Palindrome 回文相关题目汇总 LeetCode: Palindrome Partitioning 解题报告 LeetCode: Palindrome Partitioning II 解题报告 Leetcode:[DP]Longest Palindromic Substring 解题报告 LeetCode: Valid Palindrome 解题报告…

234. Palindrome Linked List[easy] Given a singly linked list, determine if it is a palindrome. Follow up:Could you do it in O(n) time and O(1) space? 解法一: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * Lis…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up: Can you solve it without using extra space? 思路:由[Leetcode]Linked List Cycle可知.利用一快一慢两个指针可以推断出链表是否存在环路. 如果两个指针相遇之前slow走了s步,则fast走了2s步.而且fast已经在长…

Given a linked list, determine if it has a cycle in it. Follow up:Can you solve it without using extra space? 给定一个链表,判断是否有环存在.Follow up: 不使用额外空间. 解法:双指针,一个慢指针每次走1步,一个快指针每次走2步的,如果有环的话,两个指针肯定会相遇. Java: public class Solution { public boolean hasCycle(Li…

Given a linked list, return the node where the cycle begins. If there is no cycle, return null. Follow up:Can you solve it without using extra space? 141. Linked List Cycle 的拓展,这题要返回环开始的节点,如果没有环返回null. 解法:双指针,还是用快慢两个指针,相遇时记下节点.参考:willduan的博客 Java: pu…

Given a singly linked list, determine if it is a palindrome. Follow up:Could you do it in O(n) time and O(1) space? Subscribe to see which companies asked this question   思路: 1.利用快慢两个指针找到中间结点 2.从中间结点下一个结点开始逆置链表,得到新的链表 3.新的链表与原来链表依次比较结点,如有不同,return fa…

Given a singly linked list, determine if it is a palindrome. Follow up:Could you do it in O(n) time and O(1) space? 题目标签:Linked List 题目给了我们一个 linked list,让我们判断它是不是回文. 这里可以利用 #206 Reverse Linked List 把右边一半的链表 倒转,然后从左右两头开始比较链表是否是回文. 这样的话,首先要找到链表的中间点,然后…

​ 请判断一个链表是否为回文链表. Given a singly linked list, determine if it is a palindrome. 示例 1: 输入: 1->2 输出: false 示例 2: 输入: 1->2->2->1 输出: true 进阶: 你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题? Follow up: Could you do it in O(n) time and O(1) space? 解题思路: 首先是寻找链表中间节…

Given a singly linked list, determine if it is a palindrome. Example 1: Input: 1->2 Output: false Example 2: Input: 1->2->2->1 Output: true Follow up:Could you do it in O(n) time and O(1) space? 这道题让我们判断一个链表是否为回文链表,LeetCode 中关于回文串的题共有六道,除了这道,其…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/palindrome-linked-list/#/description 题目描述 Given a singly linked list, determine if it is a palindrome. Follow up: Could you do it in O(n)…

Given a singly linked list, determine if it is a palindrome. Follow up:Could you do it in O(n) time and O(1) space? 思想:转置后半段链表节点,然后比较前半段和后半段节点的值是否相等. 代码如下: /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next;…

Given a singly linked list, determine if it is a palindrome. Follow up:Could you do it in O(n) time and O(1) space? 分析:题意为判断单链表是否为回文的. 思路:首先想到的是 遍历一次单链表,将其元素装入vector,然后进行第二次遍历比较来判断回文. /** * Definition for singly-linked list. * struct ListNode { * int…

Given a singly linked list, determine if it is a palindrome. 思路: 用快慢指针找到链表中点,反转后半部分链表,然后与前半部分进行匹配,随后将链表恢复原状(本题没有这个要求,具体情况具体对待). C++: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x),…

题目描述: Given a singly linked list, determine if it is a palindrome. Follow up:Could you do it in O(n) time and O(1) space? 解题思路: 使用O(n)的时间复杂度及O(1)的时间复杂度表明顺序遍历链表以及不能够开辟跟链表相当的空间,这样可以找出中间的位置,然后将后半部分的链表反转,然后跟前半部分的链表逐个位置比对. 代码如下: /** * Definition for singl…

Given a singly linked list, determine if it is a palindrome. 思路: 回文结构从后向前遍历与从前向后遍历的结果是相同的,可以利用一个栈的结构,将出栈元素与正向移动的指针指向元素比较,即可判断. 解法: /* public class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } */ import java.util.Stack; public cla…